Physics Elasticity Question 55
Question: A brass rod of cross-sectional area $ 1,cm^{2} $ and length 0.2 m is compressed lengthwise by a weight of 5 kg. If Young’s modulus of elasticity of brass is $ 1\times 10^{11},N/m^{2} $ and $ g=10,m/{{\sec }^{2}} $ , then increase in the energy of the rod will be
[MP PMT 1991]
Options:
A) $ {{10}^{-5}} $ J
B) $ 2.5\times {{10}^{-5}} $ J
C) $ 5\times {{10}^{-5}} $ J
D) $ 2.5\times {{10}^{-4}} $ J
Show Answer
Answer:
Correct Answer: B
Solution:
$ U=\frac{1}{2}\times \frac{{{\text{(stress)}}^{r}}}{Y}\times r $ = $ \frac{1}{2}\times \frac{F^{2}\times A\times L}{A^{2}\times Y} $ =$ \frac{1}{2}\times \frac{F^{2}L}{AY}=\frac{1}{2}\times \frac{{{(50)}^{2}}\times 0.2}{1\times {{10}^{-4}}\times 1\times 10^{11}} $ = $ 2.5\times {{10}^{-5}}J $
BETA
